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Here is the proposition I have:

Let $f: A \to B$ be a ring homomorphism, then $\operatorname{ker}f$ is an ideal of $A,$ and $\operatorname{Im}f$ is a subring of $B.$ And we have $A/\operatorname{ker}f \cong \operatorname{Im}f.$

And here are the definitions of the operations $+$ and $.$ in $ \langle A/\mathfrak{a}, +, . \rangle :$

If $x \in A,$ then the cosets of $A/\mathfrak{a}$ are of the form $ x + \mathfrak{a}.$ Now, define, $$ (x + \mathfrak{a}) + (y + \mathfrak{a}) = (x + y) + \mathfrak{a}$$ And $$ (x + \mathfrak{a})(y + \mathfrak{a}) = xy + \mathfrak{a}.$$ Now, I want to prove that $ \langle A/\mathfrak{a}, + \rangle$ is an abelian group in a smart way,

1- Can I say that it is an abelian group by the first Isomorphism theorem for groups? also,

2- Is the ordinary way of proving it is just by showing closure and associativity of $+$ and the existence of identity and inverse for every element in $A/\mathfrak{a}$?

Could anyone help me in answering those questions please?

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  • $\begingroup$ If $G$ is an abelian group, and $H$ is a subgroup of $G$, then $G/H = \{g+H:g \in G\}$ is also an abelian group. $\endgroup$
    – azif00
    Oct 2 at 0:02
  • $\begingroup$ @azif00 Can you please mention the theorem that says this? $\endgroup$
    – Secretly
    Nov 7 at 9:58

1 Answer 1

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How about using the first isomorphism theorem for rings? And the fact that $(A/\mathfrak a,+)$ is the abelian group of the quotient ring $(A/\mathfrak a,+,\cdot) $?

The answer to $2)$ is clearly yes. That's I guess what I should say is any way of doing is going to be equivalent to that.

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  • $\begingroup$ I do not see how I can use the first Isomorphism thm for rings? and how is this different from using the first isomorphism theorem for groups. Could you clarify the details please? $\endgroup$
    – Secretly
    Oct 2 at 0:22
  • $\begingroup$ Well the one for rings is, to my memory, completely analogous to the one for groups. Here the kernel is an ideal, and the image is a ring isomorphic to the quotient of the domain ring by that ideal. $\endgroup$
    – ACME
    Oct 2 at 0:25
  • $\begingroup$ Should not I show that the image is a ring to be able to use it? $\endgroup$
    – Secretly
    Oct 2 at 0:30
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    $\begingroup$ And, part of the definition of a ring is that it is an abelian group. $\endgroup$
    – ACME
    Oct 2 at 0:33
  • $\begingroup$ No, the image is a ring by the theorem. $\endgroup$
    – ACME
    Oct 2 at 0:34

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